Asymptotic Analysis question.

  • #1
MathematicalPhysicist
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I have this question (it's not a coursework question, I'm learning it by my own):
Use an appropiate power series expansion to find an asymptotic approximation as [tex]\epsilon[/tex] approaches 0+, correct to O(epsilon^2), for the two small roots of the equation: [tex]\epsilon x^3+x^2+2x-3=0[/tex] Then by using a suitable rescaling, find the first three terms of an asymptotic expansion as epsilon approaches 0+ of the singular root.


My way to go around this is to write down the next expansion series (because of the degree of the equation):
[tex]x(\epsilon)=\frac{1}{\epsilon^3}x_{-3}+ \frac{1}{\epsilon^2}x_{-2}+ \frac{1}{\epsilon^1}x_{-1}+x_0+x1\epsilon+...[/tex], the singular root is when epsilon equals null, i.e to solve x^2+2x-3=0=(x+3)(x-1) the roots are: x=-3 and x=1, I guess I need to choose the second root, now the recursive equation is:
[tex]x^2_{n+1}=-x_{n}/\epsilon-2/\epsilon+3/(x_{n}\epsilon)[/tex] Now, x0=1 and then use the iterative equation in order to find x_-3 and x_-2, am I correct or totally way off here?

Thanks in advance.
 

Answers and Replies

  • #3
lurflurf
Homework Helper
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h*x^3+x^2+2x-3=0
h=-1/x-2/x^2+3/x^3
take the leading term
h=-1/x
x=-1/h
x=-1/h+2+ax+bx^2+...
find more terms
 
  • #4
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I don't quite understand what your method is... but if I were to tackle this problem, i'll get rid of the messy ϵ in x^3, and write:
[tex]g(x)=f\left(\frac{1}{x}\right)=\frac{1}{x^3}(-3x^3+2x^2+x+\epsilon)[/tex]
(the motivation is that as ϵ goes to 0, the extra root goes to infinity. This is not good. So plugging in 1/x makes the root go to zero instead).


so, instead of finding roots of f, I find roots of g (much easier)
g=0 implies
[tex]x(3x^2-2x-1)=\epsilon[/tex]
i'll just expand around -1/3, 1/2, 0 and solve this eq order by order.

at the end, just take the inverse and expand using geometric series.
 

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